Handling vagueness in logic, via algebras and games ...

Handling vagueness in logic,via algebras and games.

Lecture 5.

Serafina Lapenta and Diego Valota

S. Lapenta and D. Valota (ESSLLI 2018) Lecture 5 1/34

Previously on these lectures...

... we have seen games between two players I and You, where I defends alogical formula ϕ and You attacks ϕ.

A game of this type, is appropriate for a logic L when ϕ is valid in L ifand only if I have a winning strategy.

For our last lecture, we see two games that are interesting besides anylogical interpretation:

I Ulam-Rény Game;I Akinator Game.

Previously on these lectures...

... we have seen games between two players I and You, where I defends alogical formula ϕ and You attacks ϕ.

A game of this type, is appropriate for a logic L when ϕ is valid in L ifand only if I have a winning strategy.

For our last lecture, we see two games that are interesting besides anylogical interpretation:

I Ulam-Rény Game;I Akinator Game.

Ulam-Rény GameSomeone thinks of a number between one a one million (which is just less tha 220). Anotherperson is allowed to ask up to twenty questions, to each of which the first person issupposed to answer only yes or no. Obviusly the number can be guessed by asking first: Isthe number in the first half-million? and then again reduce the reservoir of numbers in thenext question by one-half, and so on. Finally, the number is obtained in less thanlog2(1000000). Now suppose one were allowed to lie once or twice, then how manyquestions would one need to get the right answer? One clearly needs more than n questionsfor guessing one of the 2n objects because one does not know when the lie was told. Thisproblem is not solved in general.

Stanislav Ulam. Adventure of a Mathematician, 1976.

Assume that the number of questions which can be asked to figure out the ”something“ beingthought of is fixed and the one who answers is allowed to lie a certain number of times. Thequestioner, of course, doesn’t know which answer is true and which is not. Moreover the oneanswering is not required to lie as many time as is allowed....If two or more lies are allowed, then the calculation of the minimum number of questions isquite complicated... It does seem to be a very profound problem...

Alfréd Rényi. A Diary on Information Theory, 1976.

Ulam-Rény GameSomeone thinks of a number between one a one million (which is just less tha 220). Anotherperson is allowed to ask up to twenty questions, to each of which the first person issupposed to answer only yes or no. Obviusly the number can be guessed by asking first: Isthe number in the first half-million? and then again reduce the reservoir of numbers in thenext question by one-half, and so on. Finally, the number is obtained in less thanlog2(1000000). Now suppose one were allowed to lie once or twice, then how manyquestions would one need to get the right answer? One clearly needs more than n questionsfor guessing one of the 2n objects because one does not know when the lie was told. Thisproblem is not solved in general.

Stanislav Ulam. Adventure of a Mathematician, 1976.

Assume that the number of questions which can be asked to figure out the ”something“ beingthought of is fixed and the one who answers is allowed to lie a certain number of times. Thequestioner, of course, doesn’t know which answer is true and which is not. Moreover the oneanswering is not required to lie as many time as is allowed....If two or more lies are allowed, then the calculation of the minimum number of questions isquite complicated... It does seem to be a very profound problem...

Alfréd Rényi. A Diary on Information Theory, 1976.

Ulam-Rény Game

I Two players: Questioner Q and Responder R;I A search space S = 1, . . . , n, n ∈ N+;I R chooses x ∈ S ;I to identify x , Q asks YES-NO questions to R;

I ... a.k.a. Twenty Questions game.

Ulam-Rény Game

I Two players: Questioner Q and Responder R;I A search space S = 1, . . . , n, n ∈ N+;I R chooses x ∈ S ;I to identify x , Q asks YES-NO questions to R;I ... a.k.a. Twenty Questions game.

Denote by Ω ⊆ S the set of numbers candidates to be x because of theanswers already received.

Naive Strategy:I Q: ”Is x = 1?“; R: ”YES“, End;I R: ”NO“, then;

I Q: ”Is x = 2?“; R: ”YES“, End;I R: ”NO“, then;

...I Q: ”Is x = n?“; R: ”YES“,

Worst case: n questions.

Smart Strategy:Start with Ω = S

1. partition Ω in two sets X1 andX2 such that ||X1| − |X2|| ≤ 1;

2. Q: ”Is x ∈ X1?“;3. R: ”YES“, then Ω = X1;4. R: ”NO“, then Ω = X2;5. back to 1.;

Worst case: dlog2(n)e questions.

Denote by Ω ⊆ S the set of numbers candidates to be x because of theanswers already received.

Naive Strategy:I Q: ”Is x = 1?“; R: ”YES“, End;I R: ”NO“, then;

I Q: ”Is x = 2?“; R: ”YES“, End;I R: ”NO“, then;

...I Q: ”Is x = n?“; R: ”YES“,

Worst case: n questions.

Smart Strategy:Start with Ω = S

1. partition Ω in two sets X1 andX2 such that ||X1| − |X2|| ≤ 1;

2. Q: ”Is x ∈ X1?“;3. R: ”YES“, then Ω = X1;4. R: ”NO“, then Ω = X2;5. back to 1.;

Worst case: dlog2(n)e questions.

We identify a question Q with the subset Q ⊆ S of numbers for which theanswer to Q is YES.

Q ask Q and R answers YES, then each x ∈ Q satisfies the answer to Q ,and each x 6∈ Q falsifies the answer.

Q ask Q and R answers NO, then each x 6∈ Q satisfies the answer to Q ,and each x ∈ Q falsifies the answer.

x satisfies the answer YES to Q iff satisfies the answer NO to S \ Q ,x falsifies the answer NO to Q iff falsifies the answer YES to S \ Q .

Since any elements of Ω can be x , and any other element cannot. Then, astate of the game is completely determined by Ω:

s(x) =

1 if x ∈ Ω

0 otherwise

s(x) =

1 if x ∈ Ω

0 otherwise

s(x) =

1 if x ∈ Ω

0 otherwise

s(x) =

1 if x ∈ Ω

0 otherwise

Let s1 and s2 be two states. We say that a s1 is at least as informative asa s2, denoted by s1 ≤ s2, if all the numbers excluded by s2 are alsoexcluded by s1.

The state 1 constantly 1 is the the least informative one, corresponds atsaying x ∈ S .The state 0 constantly 0 is the the most informative one, and itcorresponds to the inconsistent information.

The minimal information of both s1 and s2 is the function(s1 ∧ s2)(x) = mins1(x), s2(x).The state representing the smallest amount of information which added tos1 yields a refinement of s2 is the implication s1 → s2 of the states.

The set of formulas ϕ(p1, . . . , pn) such that for any choice of statess1, . . . , sn , we have ϕ(s1, . . . , sn) = 1 is classical logic.

Ulam-Rény Game with LiesSuppose that R can lies at most e times, for a fixed and known e ≥ 0.

A number y is entitled to be x if it satisfies all the answers.Additionally, y cannot be x if it falsifies at least e + 1 answers.

Finally, y may falsify i questions for each i = 1, . . . , e , and still be acandidate to be x .

For each number y there exists e + 1 states where y may be x , and onlyone where it is not.

The state function s(x) : S → 0, 1e+1 , . . . ,

ee+1 , 1 is defined by:

s(x) =

e+1−ie+1 iff x falsifies i answers,

0 iff x falsifies ≥ e + 1 answers.

s(x) =

Q: Is the number even? R: YES Q: Is the number even? R: NO

The state is not inconsistent ⇒ The number of lies is decreased by one!

Now suppose e = 1.Q: Is the number even? R: YES Q: Is the number even? R: YES

⇒ Since e=1, Q is sure that x is even.

We conclude that the conjunction of two (opposite) answers is notcontractive, a 6= a&a.

States are functions s : S → 0, 1e+1 , . . . ,

ee+1 , 1.

The information carried by a question Q and its answer a, is representedby a function f(Q,a) : S → e

e+1 , 1 given by

f(Q,a)(x) = 1 if (x ∈ Q and a = YES ) or (x 6∈ Q and a = NO )

f(Q,a)(x) =e

e + 1if (x 6∈ Q and a = YES ) or (x ∈ Q and a = NO )

Let s be the state before Q . Then, for all x ∈ S we compute the next states ′ by

s ′ = s f(Q,a) s ′(x) = s(x) f(Q,a)(x).

Where x y = max0, x + y − 1 is extended to functions by(f g)(x) = f (x) g(x).

States are functions s : S → 0, 1e+1 , . . . ,

ee+1 , 1.

The information carried by a question Q and its answer a, is representedby a function f(Q,a) : S → e

e+1 , 1 given by

f(Q,a)(x) = 1 if (x ∈ Q and a = YES ) or (x 6∈ Q and a = NO )

f(Q,a)(x) =e

e + 1if (x 6∈ Q and a = YES ) or (x ∈ Q and a = NO )

Let s be the state before Q . Then, for all x ∈ S we compute the next states ′ by

s ′ = s f(Q,a) s ′(x) = s(x) f(Q,a)(x).

Where x y = max0, x + y − 1 is extended to functions by(f g)(x) = f (x) g(x).

Let In = 0, 1n , . . . ,

n−1n , 1.

Every Łukasiewicz logic formula ϕ(x1, . . . , xm) represents a functionsfϕ : (In)m → In .

Let C (n, e) =

1e+1 ,

Theorem (Mundici, 1992)For each function f : C (n, e)→ Ie+2, there is a formula ϕ such that fequals the restriction of fϕ to C (n, e).For each formula ϕ, the restriction of fϕ to C (n, e), is a functions : C (n, e)→ 0, 1

e+2 , . . . ,e+1e+2 , 1.

Let In = 0, 1n , . . . ,

n−1n , 1.

Let C (n, e) =

1e+1 ,

e+2 , . . . ,e+1e+2 , 1.

Let In = 0, 1n , . . . ,

n−1n , 1.

Let C (n, e) =

1e+1 ,

e+2 , . . . ,e+1e+2 , 1.

Ulam-Rény Game Łukasiewicz logicmax e of lies truth-values 0, 1

e+1 , . . . ,e

e+1 , 1search space S = 0, 1n C (n, e) = 1

e+1 ,e

x ∈ S point in C (n, e)

initial state of knowledge a tautologyfinal state of knowledge ϕ uniquely sat in C (n, e)

current state of knoledge s formula ϕ s.t. fϕ = s in C (n, e)

question Q subset of C (n, e)

positive answer ψ to Q formula ϕ s.t. fψ = fϕ in C (n, e)

state s after answer to Q1, . . .Qi conjunction of corr. formulasset of excluded numbers points of C (n, e) falsifying ϕ

Other worksI Changing the setting of the game:

R answers to Q now travel on a noisy channel, and lies becamestransmission errors;

Ulam-Rény Games are used to study error-correcting codes.

F. CicaleseFault tolerant search algorithms.Springer, Berlin (2013).

I Increasing the number of transmission channels, we obtainUlam-Rény Games for BL, Gödel and Product logic.

E.A. Corsi, F. Montagna.The Rényi-Ulam games and many-valued logics.Fuzzy Sets and Systems 301: 37-50 (2016)

And now, from Liers and Errors to Genii!!

R answers to Q now travel on a noisy channel, and lies becamestransmission errors;Ulam-Rény Games are used to study error-correcting codes.

And now, from Liers and Errors to Genii!!S. Lapenta and D. Valota (ESSLLI 2018) Lecture 5 13/34

Game semantics for Łukasiewicz-Moisil logic

Diaconescu and Leuştean’ model for LM-logicLet C be a set of characters.

The game:Player: thinks of a character c ∈ C

Genie: asks questionsPlayer: is allowed to answer with:

Yes, Probably, Don’t know, Probably not, Not.

A set of characters is displayed after n questions.

A remarkIn the actual game, after a number of questions, the Genie either displaysa character and asks if his guess is correct, or it displays a set ofcharacters and asks if the Player’s choice is in the list.

Diaconescu and Leuştean’ model for LM-logicLet C be a set of characters.

The game:Player: thinks of a character c ∈ C

Genie: asks questionsPlayer: is allowed to answer with:

Yes, Probably, Don’t know, Probably not, Not.

A set of characters is displayed after n questions.

A remarkIn the actual game, after a number of questions, the Genie either displaysa character and asks if his guess is correct, or it displays a set ofcharacters and asks if the Player’s choice is in the list.

Akinator with n questions and Yes/No answers

The Genie has a data base of characters:

Hannibal Don Vito Princess Jean-Luc CruellaLector Corleone Leia Picard de Vil

We have a set C of characters.

Akinator with n questions and Yes/No answersThe Genie has a description of the characters:

villain 1 1 0 0 1

eccentric hair 0 0 1 1 1

bossy 1 1 1 1 1

european 1 1 0 1 1

The Genie asks questions of the form Is your character bossy?

The Genie’s description of the characters is a set ∆ ⊆ LC2 .A question is an element δ ∈ ∆.

villain 1 1 0 0 1

bossy 1 1 1 1 1

european 1 1 0 1 1

villain 1 1 0 0 1

bossy 1 1 1 1 1

european 1 1 0 1 1

villain 1 1 0 0 1

bossy 1 1 1 1 1

european 1 1 0 1 1

villain 1 1 0 0 1

bossy 1 1 1 1 1

european 1 1 0 1 1

villain 1 1 0 0 1

bossy 1 1 1 1 1

european 1 1 0 1 1

villain 1 1 0 0 1

bossy 1 1 1 1 1

european 1 1 0 1 1

villain 1 1 0 0 1

bossy 1 1 1 1 1

european 1 1 0 1 1

A round is a pair (question,answer).

A round is represented as Qaδ where δ ∈ ∆ and a ∈ L2.

I Is your character a villain? No.I Q0

villain

A game is a sequence of n rounds γ := Qa1δ1. . .Qan

villain

The knowledge after a round

Q0villain is

villain 1 1 0 0 1K 0

villain 0 0 1 1 0

The knowledge after a round Qaδ is K a

δ = 1δ−1(a) ∈ LC2

where 1X is the characteristic function of X in C .

The knowledge after a round Q0villain is

villain 1 1 0 0 1

K 0villain 0 0 1 1 0

δ = 1δ−1(a) ∈ LC2

villain 0 0 1 1 0

δ = 1δ−1(a) ∈ LC2

villain 0 0 1 1 0

δ = 1δ−1(a) ∈ LC2

Akinator with n questions and Yes/No answersSuppose we have two rounds Q1

european and Q0villain.

The knowledge is

K 1european 1 1 0 1 1

After the two rounds, we choose Jean Luc Picard.At each round k we select a set of characters Ck ⊆ C such that

1Ck= K a1

δ1∧ . . . ∧ K ak

Where are the nuances?

european and Q0villain. The knowledge is

1Ck= K a1

δ1∧ . . . ∧ K ak

After the two rounds, we choose Jean Luc Picard.

At each round k we select a set of characters Ck ⊆ C such that

1Ck= K a1

δ1∧ . . . ∧ K ak

1Ck= K a1

δ1∧ . . . ∧ K ak

1Ck= K a1

δ1∧ . . . ∧ K ak

Recalling the adjunction from Lecture 3

If n ≥ 1 and B is a Boolean algebra then

Dn(B) = 〈x1, . . . , xn〉 ∈ Bn | x1 ≥ . . . ≥ xn

has a natural structure of n-nuanced Łukasiewicz-Moisil algebra:

I the lattice operations are defined pointwiseI ¬〈x1, . . . , xn〉 = (¬xn, . . . ,¬x1)

I ϕi 〈x1, . . . , xn〉 = 〈xi , . . . , xi 〉 for any i ∈ [n]

Towards Moisil logicWe interpret a k th round Qa

Ik(Qaδ ) = 〈1, . . . , 1︸︷︷︸

,K aδ , . . . ,K

aδ︸︷︷︸

n−k+1

〉 ∈ (LC2 )n

We interpret a sequence of rounds by

I(Qa1δ1

) = I1(Qa1δ1

I(Qa1δ1. . .Qak

δkQak+1δk+1

) = I(Qa1δ1. . .Qak

δk) ∧ Ik+1(Qak+1

δk+1)

Note that for a game γ = Qa1δ1. . .Qan

δnwe have

I(γ) = 〈1C1 , . . . , 1Cn〉 ∈ (LC2 )n

Obviously, the answer of the Genie is Cn .

Ik(Qaδ ) = 〈1, . . . , 1︸︷︷︸

,K aδ , . . . ,K

aδ︸︷︷︸

n−k+1

〉 ∈ (LC2 )n

I(Qa1δ1

) = I1(Qa1δ1

I(Qa1δ1. . .Qak

δkQak+1δk+1

) = I(Qa1δ1. . .Qak

δk) ∧ Ik+1(Qak+1

δk+1)

δnwe have

I(γ) = 〈1C1 , . . . , 1Cn〉 ∈ (LC2 )n

Ik(Qaδ ) = 〈1, . . . , 1︸︷︷︸

,K aδ , . . . ,K

aδ︸︷︷︸

n−k+1

〉 ∈ (LC2 )n

I(Qa1δ1

) = I1(Qa1δ1

I(Qa1δ1. . .Qak

δkQak+1δk+1

) = I(Qa1δ1. . .Qak

δk) ∧ Ik+1(Qak+1

δk+1)

δnwe have

I(γ) = 〈1C1 , . . . , 1Cn〉 ∈ (LC2 )n

Ik(Qaδ ) = 〈1, . . . , 1︸︷︷︸

,K aδ , . . . ,K

aδ︸︷︷︸

n−k+1

〉 ∈ (LC2 )n

I(Qa1δ1

) = I1(Qa1δ1

I(Qa1δ1. . .Qak

δkQak+1δk+1

) = I(Qa1δ1. . .Qak

δk) ∧ Ik+1(Qak+1

δk+1)

δnwe have

I(γ) = 〈1C1 , . . . , 1Cn〉 ∈ (LC2 )n

Ik(Qaδ ) = 〈1, . . . , 1︸︷︷︸

,K aδ , . . . ,K

aδ︸︷︷︸

n−k+1

〉 ∈ (LC2 )n

I(Qa1δ1

) = I1(Qa1δ1

I(Qa1δ1. . .Qak

δkQak+1δk+1

) = I(Qa1δ1. . .Qak

δk) ∧ Ik+1(Qak+1

δk+1)

δnwe have

I(γ) = 〈1C1 , . . . , 1Cn〉 ∈ (LC2 )n

Obviously, the answer of the Genie is Cn .S. Lapenta and D. Valota (ESSLLI 2018) Lecture 5 23/34

Towards Moisil logicThe following map is an isomorphism

ι : Dn(L2)→ Ln+1, ι(〈1 . . . , 1︸︷︷︸k

, 0, . . . , 0〉) =k

We have the isomorphism

ιC : Tn(LC2 )→ LCn+1, ιC (〈f1, . . . , fn〉)(c) = ι(〈f1(c), . . . , fn(c)〉).

For a game γ we have the interpretation I(γ) ∈ Tn(LC2 ). We set

V(γ) : C → Ln+1, V(γ) = ιC (I(γ)).

What is the meaning of V(γ)?

ι : Dn(L2)→ Ln+1, ι(〈1 . . . , 1︸︷︷︸k

, 0, . . . , 0〉) =k

V(γ) : C → Ln+1, V(γ) = ιC (I(γ)).

ι : Dn(L2)→ Ln+1, ι(〈1 . . . , 1︸︷︷︸k

, 0, . . . , 0〉) =k

For a game γ we have the interpretation I(γ) ∈ Tn(LC2 ).

We set

V(γ) : C → Ln+1, V(γ) = ιC (I(γ)).

ι : Dn(L2)→ Ln+1, ι(〈1 . . . , 1︸︷︷︸k

, 0, . . . , 0〉) =k

V(γ) : C → Ln+1, V(γ) = ιC (I(γ)).

ι : Dn(L2)→ Ln+1, ι(〈1 . . . , 1︸︷︷︸k

, 0, . . . , 0〉) =k

V(γ) : C → Ln+1, V(γ) = ιC (I(γ)).

Akinator with 3 questions and Yes/No answersγ = Q1

eccentric hair Q0villain Q1

european

K 1eccentric hair 0 0 1 1 1

We have the following intermediate sets of characters

I C1 = Princess Leia, Jean-Luc Picard, Cruella de Vil

I C2 = Princess Leia, Jean-Luc Picard

I C3 = Jean-Luc Picard

Akinator with 3 questions and Yes/No answersγ = Q1

eccentric hair Q0villain Q1

european

K 1eccentric hair 0 0 1 1 1

We have the following intermediate sets of characters

I C1 = Princess Leia, Jean-Luc Picard, Cruella de Vil

I C2 = Princess Leia, Jean-Luc Picard

I C3 = Jean-Luc Picard

Akinator with 3 questions and Yes/No answers

We have the following intermediate sets of characters:I C1 = Princess Leia, Jean-Luc Picard, Cruella de VilI C2 = Princess Leia, Jean-Luc PicardI C3 = Jean-Luc Picard

V(γ)(c) = ιC (I(γ))(c) = ι(〈1C1(c), 1C2(c), 1C3(c)〉)

I V(γ)(Hannibal Lector) = ι(〈0, 0, 0〉) = 0

I V(γ)(Princess Leia) = ι(〈1, 1, 0〉) = 23

I V(γ)(Jean-Luc Picard) = ι(〈1, 1, 1〉) = 1

Towards Moisil logic

If we consider C0 = C , for any 0 ≤ k < n we have

V(γ)(c) = kn iff c ∈ Ck \ Ck+1

the character c was eliminated by the (k+1)-th question.

Moreover,

V(γ)(c) = 1 iff c ∈ Cn .

It is a temporal interpretation of the Boolean nuances!

If we consider C0 = C , for any 0 ≤ k < n we have

V(γ)(c) = kn iff c ∈ Ck \ Ck+1

the character c was eliminated by the (k+1)-th question.

Moreover,

V(γ)(c) = 1 iff c ∈ Cn .

It is a temporal interpretation of the Boolean nuances!

We define

G (C ,∆, n) = V(γ) : C → Ln+1 | γ is a game.

In general, G (C ,∆, n) is not an Łukasiewicz-Moisil algebra.

Let Ak(C ,∆, n) be the Łukasiewicz-Moisil algebra generated by

G (C ,∆, n) in Tn(LC2 ).

If any set of characters is selected by a question (i.e., ∆ = LC2 ) then

Ak(C ,∆, n) = G (C ,∆, n) = Tn(LC2 ) ' LCn+1.

We define

G (C ,∆, n) in Tn(LC2 ).

Ak(C ,∆, n) = G (C ,∆, n) = Tn(LC2 ) ' LCn+1.

We define

G (C ,∆, n) in Tn(LC2 ).

Ak(C ,∆, n) = G (C ,∆, n) = Tn(LC2 ) ' LCn+1.

We define

G (C ,∆, n) in Tn(LC2 ).

Ak(C ,∆, n) = G (C ,∆, n) = Tn(LC2 ) ' LCn+1.

Moisil logic

Theorem

The following are equivalent:I A is derivable in n-nuanced Moisil logic.I e(A) = 1 for any C , ∆, and evaluation e in Ak(C ,∆, n).

Akinator with n questions and p answers

The set of answers is a partiallyordered set (Ans,≺).

Don′t know ≺ Probably ≺ Yes

Don′t know ≺ Probably not ≺ No .

Akinator with n questions and p answersI Answers: an ordered set 〈Lp,〉 with p ≥ 2

I Characters: a set CI Description of the characters: ∆ ⊆ LCp

I A round (question,answer): Qδa with δ ∈ ∆ and a ∈ Lp

I Knowledge after a round Qaδ :

K aδ =

∨a x

1δ−1(x) ∈ LC2

Even if there are more than two possible answers,the selection is a (classical) set of characters.

I A game: γ = Qa1δ1. . .Qan

I We define I(γ) ∈ Tn(LC2 ) and V(γ) ∈ LCn+1 as before

K aδ =

∨a x

1δ−1(x) ∈ LC2

K aδ =

∨a x

1δ−1(x) ∈ LC2

Classical Akinator with n questions and lies

It is a work in progress by D. Diaconescu and I. Leuştean.

Classical Akinator with n questions and 1 lie

I The set of answers is L2.I The player is allowed to lie at most 1 time.I C is the set of characters, D ⊆ L2

C is the set of data.I We assume 1, 0 ∈ D .I (question,answer): Qa

δ with δ ∈ D and a ∈ L2,but the answer might be a lie.

I The knowledge after Qaδ is K a

δ ∈ LC3 , where

K 1δ = δ ⊕ ( 1

2 ¬δ), K 0δ = ¬δ ⊕ ( 1

2 δ), with Łukasiewicz ⊕ and .I A game is γ := Qa1

δ1· · · · · ·Qan

It will give a game semantics for the theory of Dn(L3).

Classical Akinator with n questions and 1 lie

I The set of answers is L2.I The player is allowed to lie at most 1 time.I C is the set of characters, D ⊆ L2

C is the set of data.I We assume 1, 0 ∈ D .I (question,answer): Qa

δ with δ ∈ D and a ∈ L2,but the answer might be a lie.

I The knowledge after Qaδ is K a

δ ∈ LC3 , where

K 1δ = δ ⊕ ( 1

2 ¬δ), K 0δ = ¬δ ⊕ ( 1

2 δ), with Łukasiewicz ⊕ and .I A game is γ := Qa1

δ1· · · · · ·Qan

It will give a game semantics for the theory of Dn(L3).

Thank you!!

Handling vagueness in logic, via algebras and games ...

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